What are the extrema of  f(x)=x/(x-2) on the interval [-5,5]?

Jan 17, 2016

Answer:

There are no absolute extrema, and the existence of relative extrema depends on your definition of relative extrema.

Explanation:

$f \left(x\right) = \frac{x}{x - 2}$ increases without bound as $x \rightarrow 2$ from the right.
That is: ${\lim}_{x \rightarrow {2}^{+}} f \left(x\right) = \infty$
So, the function has no absolute maximum on $\left[- 5 , 5\right]$

$f$ decreases without bound as $x \rightarrow 2$ from the left, so there is no absolute minimum on $\left[- 5 , 5\right]$.

Now, $f ' \left(x\right) = \frac{- 2}{x - 2} ^ 2$ is always negative, so, taking the domain to be $\left[- 5 , 2\right) \cup \left(2 , 5\right]$ , the function decreases on $\left[- 5 , 2\right)$ and on $\left(2 , 5\right]$.

This tells us that $f \left(- 5\right)$ is the greatest value of $f$ nearby considering only $x$ values in the domain. It is a one-sided relative maximum. Not all treatments of calculus allow one sided relative extrema.

Similarly, if your approach allow one-sided relative extrema, then #f(5) is a relative mimimum.

To help visualize, here is a graph. The restricted domain graph is solid and the endpoints are marked.

The natural domain graph extends into the dashed line part of the picture.